SOLUTION: Given the function f(x)=8-x^3, x>_0 (line goes under >) a) Sketch the graph of f (use a solid line, label f(x) b) Use the graph of f above and sketch the graph of f^-1 on the sam

Algebra ->  Trigonometry-basics -> SOLUTION: Given the function f(x)=8-x^3, x>_0 (line goes under >) a) Sketch the graph of f (use a solid line, label f(x) b) Use the graph of f above and sketch the graph of f^-1 on the sam      Log On


   

Question 1190910: Given the function f(x)=8-x^3, x>_0 (line goes under >)
a) Sketch the graph of f (use a solid line, label f(x)
b) Use the graph of f above and sketch the graph of f^-1 on the same axis as in part a) of this problem (use a dashed line, label the inverse). How did you find the points to sketch the inverse function?
c) Find f^-1(x)
Found 2 solutions by MathLover1, Boreal:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=8-x%5E3, x%3E=0+
a.
download


b.
desmos-graph-11
note: orange line should be dashed line
c.
inverse of f%28x%29=8-x%5E3, x%3E=0 will be:
f%28x%29=y
y=8-x%5E3 ....swap variables
x=8-y%5E3
y%5E3=8-x
y=root%283%2C8-x%29
f%5E-1%28x%29=root%283%2C8-x%29 , x%3E=0


Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
f(x)=8-x^3
inverse
x=8-y^3
y^3=8-x
y=(8-x)^(1/3)
graph%28300%2C300%2C0%2C10%2C-10%2C10%2C8-x%5E3%2C%288-x%29%5E%281%2F3%29%29
To find the points for the inverse, (8, 0) is one, and (0, 2) is the other. The two are equal when
8-x^3=(8-x)^(1/3)
One way is to graph both and find where they intersect, which is at x=1.834.
One can put in points for x and take the cube root.